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  4. In the expansion of (1+x+x2+x3)6, the coefficient of x14 is

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Q. In the expansion of $ {{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}}, $ the coefficient of $ {{x}^{14}} $ is

KEAMKEAM 2007Binomial Theorem

Solution:

$ {{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}} $
$={{(1+x)}^{6}}{{(1+{{x}^{2}})}^{6}} $
$={{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}x{{+}^{6}}{{C}_{2}}{{x}^{2}}{{+}^{6}}{{C}_{3}}{{x}^{3}}{{+}^{6}}{{C}_{4}}{{x}^{4}} $ $ {{+}^{6}}{{C}_{5}}{{x}^{5}}{{+}^{6}}{{C}_{6}}{{x}^{6}}) $ $ \times {{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}{{x}^{2}}{{+}^{6}}{{C}_{2}}{{x}^{4}}{{+}^{6}}{{C}_{3}}{{x}^{6}}{{+}^{6}}{{C}_{4}}{{x}^{8}} $ $ {{+}^{6}}{{C}_{5}}{{x}^{10}}{{+}^{6}}{{C}_{6}}{{x}^{12}}) $
$ \therefore $ Coefficient of $ {{x}^{14}} $ in $ {{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}} $ $ {{=}^{6}}{{C}_{2}}{{.}^{6}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{6}}{{C}_{5}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{4}} $
$=\frac{6!}{2!4!}.\frac{6!}{0!6!}+\frac{6!}{4!2!}.\frac{6!}{5!1!}+\frac{6!}{0!6!}.\frac{6!}{2!4!} $
$=15+90+15=120 $